1. **Stating the problem:** We want to evaluate the integral $$\int \frac{-2}{2x^2 - x^3 - x} \, dx$$. 2. **Simplify the denominator:** Factor the denominator: $$2x^2 - x^3 - x = -x (x - 1)^2$$ 3. **Rewrite the integral:** Using the factorization, the integral becomes: $$\int \frac{-2}{-x (x - 1)^2} \, dx = \int \frac{2}{x (x - 1)^2} \, dx$$ 4. **Partial fraction decomposition:** We express: $$\frac{2}{x (x - 1)^2} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2}$$ Multiply both sides by the denominator $x (x - 1)^2$: $$2 = A (x - 1)^2 + B x (x - 1) + C x$$ Expand: $$2 = A (x^2 - 2x + 1) + B (x^2 - x) + C x$$ $$= (A + B) x^2 + (-2A - B + C) x + A$$ 5. **Equate coefficients:** For the polynomial identity to hold for all $x$: - Coefficient of $x^2$: $A + B = 0$ - Coefficient of $x$: $-2A - B + C = 0$ - Constant term: $A = 2$ 6. **Solve the system:** From $A = 2$, substitute into $A + B = 0$: $$2 + B = 0 \implies B = -2$$ Substitute $A$ and $B$ into $-2A - B + C = 0$: $$-2(2) - (-2) + C = 0 \implies -4 + 2 + C = 0 \implies C = 2$$ 7. **Rewrite the integral with partial fractions:** $$\int \frac{2}{x} \, dx + \int \frac{-2}{x - 1} \, dx + \int \frac{2}{(x - 1)^2} \, dx$$ 8. **Integrate each term:** - $$\int \frac{2}{x} \, dx = 2 \ln|x| + C_1$$ - $$\int \frac{-2}{x - 1} \, dx = -2 \ln|x - 1| + C_2$$ - $$\int \frac{2}{(x - 1)^2} \, dx = 2 \int (x - 1)^{-2} \, dx = 2 \cdot \frac{(x - 1)^{-1}}{-1} + C_3 = -\frac{2}{x - 1} + C_3$$ 9. **Combine results:** $$\int \frac{-2}{2x^2 - x^3 - x} \, dx = 2 \ln|x| - 2 \ln|x - 1| - \frac{2}{x - 1} + C$$ 10. **Simplify logarithms:** $$2 \ln|x| - 2 \ln|x - 1| = 2 \ln \left| \frac{x}{x - 1} \right|$$ **Final answer:** $$\boxed{\int \frac{-2}{2x^2 - x^3 - x} \, dx = 2 \ln \left| \frac{x}{x - 1} \right| - \frac{2}{x - 1} + C}$$ where $C$ is an arbitrary constant of integration.